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Chapter 8: Problem 26

Prove Lemma 8.6: Let \(E\) be an elementary matrix. Then \(|E A|=|E||A|\)

Short Answer

Expert verified
We proved the lemma |E A| = |E||A| for an elementary matrix E and another matrix A by analyzing the effect of each type of elementary row operation on the determinant: For row swapping, |E| = -1 and |E A| = |-1||A| = |E||A|; for row scaling, |E| = k and |E A| = |k||A| = |E||A|; and for row addition, |E| = 1 and |E A| = |1||A| = |E||A|. In all three cases, the lemma holds true, completing the proof.

Step by step solution

01

Recall the definition of elementary matrices

Elementary matrices are matrices that can be obtained from the identity matrix by applying a single elementary row operation. There are three types of elementary row operations: 1. Swap two rows. 2. Multiply a row by a nonzero scalar. 3. Add the multiple of one row to another row. Let's analyze the effect of each type of elementary row operation on the determinant.
02

Analyze the effect of row swapping on the determinant

When we swap two rows of a matrix, the determinant's sign is inverted. So, if we have an elementary matrix E that represents a row swap, |E| will be -1. Let's denote the matrix after swapping rows of A as A'. Then, it's known that |A'| = -|A|. Therefore, we have |E A| = |-1||A| = |E||A|.
03

Analyze the effect of row scaling on the determinant

When we multiply a row of a matrix by a nonzero scalar k, the determinant is also multiplied by the same scalar k. So, if we have an elementary matrix E that represents a row scaling, |E| will be the scalar k. Let's denote the matrix after multiplying a row of A by k as A'. Then, it's known that |A'| = k|A|. Therefore, we have |E A| = |k||A| = |E||A|.
04

Analyze the effect of row addition on the determinant

When we add the multiple of one row to another row, the determinant doesn't change. So, if we have an elementary matrix E that represents a row addition, |E| will be 1. Let's denote the matrix after adding the multiple of one row to another row of A as A'. Then, it's known that |A'| = |A|. Therefore, we have |E A| = |1||A| = |E||A|.
05

Combine the results for different elementary row operations

In all three cases of elementary row operations, we have shown that |E A| = |E||A|. Therefore, we can conclude that for any elementary matrix E, the lemma holds true: |E A| = |E||A|. This completes the proof.

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Most popular questions from this chapter

Prove Theorem 8.13: Let \(F\) and \(G\) be linear operators on a vector space \(V\). Then (i) \(\operatorname{det}(F \circ G)=\operatorname{det}(F) \operatorname{det}(G)\) (ii) \(F\) is invertible if and only if \(\operatorname{det}(F) \neq 0\)

Find the determinant of \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) defined by $$ F(x, y, z)=(x+3 y-4 z, 2 y+7 z, x+5 y-3 z) $$The determinant of a linear operator \(F\) is equal to the determinant of any matrix that represents \(F\). Thus first find the matrix \(A\) representing \(F\) in the usual basis (whose rows, respectively, consist of the coefficients of $x, y, z$ ). Then $$ A=\left[\begin{array}{rrr} 1 & 3 & -4 \\ 0 & 2 & 7 \\ 1 & 5 & -3 \end{array}\right], \quad \text { and so } \quad \operatorname{det}(F)=|A|=-6+21+0+8-35-0=-8 $$

Suppose \(B\) is row equivalent to a square matrix \(A\). Prove that \(|B|=0\) if and only if \(|A|=0\).

Let \(V\) be the space of \(m\) -square matrices (as above), and suppose \(D: V \rightarrow K\). Show that the following weaker statement is equivalent to \(D\) being alternating: $$D\left(A_{1}, A_{2}, \ldots, A_{n}\right)=0 \quad \text { whenever } \quad A_{i}=A_{i+1} \text { for some } i$$

Prove Theorem \(8.10\) (Cramer's rule): The (square) system \(A X=B\) has a unique solution if and only if \(D \neq 0 .\) In this case, \(x_{i}=N_{i} / D\) for each \(i\). By previous results, \(A X=B\) has a unique solution if and only if \(A\) is invertible, and \(A\) is invertible if and only if \(D=|A| \neq 0\) Now suppose \(D \neq 0 .\) By Theorem $8.9, A^{-1}=(1 / D)(\operatorname{adj} A) .\( Multiplying \)A X=B\( by \)A^{-1}$, we obtain $$ X=A^{-1} A X=(1 / D)(\operatorname{adj} A) B $$ Note that the ith row of \((1 / D)(\operatorname{adj} A)\) is $(1 / D)\left(A_{1 i}, A_{2 i}, \ldots, A_{n i}\right) .\( If \)B=\left(b_{1}, b_{2}, \ldots, b_{n}\right)^{T}$, then, by (1), $$ x_{i}=(1 / D)\left(b_{1} A_{1 i}+b_{2} A_{2 i}+\cdots+b_{n} A_{n i}\right) $$ However, as in Problem $8.33, b_{1} A_{1 i}+b_{2} A_{2 i}+\cdots+b_{n} A_{n i}=N_{i}$, the determinant of the matrix obtained by replacing the ith column of \(A\) by the column vector \(B .\) Thus, \(x_{i}=(1 / D) N_{i}\), as required.
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